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A body of mass 50kg starts from rest and slides down a smooth incilined plane.Find the force ..p.. applied at an angle 30^(@) to theinclined plane so that the body attains a velocity of 10ms^(-1) after sliding a distance of 7m. |
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Answer» Solution :Work done by force P in displacing the block by 7m `W_(1) (Fcostheta)s=Pxx7xxcos30^(@)=(7sqrt(3))/(2)Pj` work done by gravity `W_(2)(mg sintheta)S=50xx9.8xx(1)/(2)xx7=1715j` USING work - energy theorem, `W = DELTA K.E.` `or W_(1) +W_(2)=(1)/(2)m(v_(f)^(2)-v_(i)^(2))` `or(7sqrt(3))/(2)P+1715=(1)/(2)xx50xx(10^(2)-0^(2))` `or6.06 + 1715 = 2500 therefore P=129.5N` |
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